Chinese remainder theorem for rings applied to Z/nZ

I've been working with the Chinese Remainder Theorem for rings:

Let $R$ be a ring and let ${A}_{1}, {A}_{2}, ..., {A}_{n}$ be ideals in $R$ such that for each $i ,j \ \epsilon \ {1,2,...,n}, {A}_{i}$ is comaximal to ${A}_{j}$ for each $i \neq j$. Then $R/({A}_{1}{A}_{2}...{A}_{n}) \simeq R/{A}_{1} \times R/{A}_{2} \times ... \times R/{A}_{n}$.

I have no problem with the proof but, the book that I'm using says that this Theorem gives the following isomorphisms:

1. $(Z/nmZ)^{ \times }\simeq (Z/nZ)^{ \times }\times(Z/mZ)^{ \times }$

2. If $n=({p}_{1})^{{\alpha}_{1}} ({p}_{2})^{{\alpha}_{2}}...({p}_{k})^{{\alpha}_{k}}\$ then $Z/nZ\quad \simeq \quad Z/{ { p }_{ 1 } }^{ { \alpha }_{ 1 } }Z\quad \times \quad Z/{ { p }_{ 2 } }^{ { \alpha }_{ 2 } }Z\quad \times \quad ...\quad \times \quad Z/{ { p }_{ k } }^{ { \alpha }_{ k } }Z$

I have no idea how to prove this with the Chinese Remainder Theorem. Isn't it necessary that $Z/{ p }_{ i } ^{ { \alpha }_{ i }} Z$ is ideal, and comaximal to the others $Z/{ p }_{ j} ^{ { \alpha }_{ j }} Z$? How can i prove this?

• notice: the ideals of the CRT are $p_i^{\alpha_i}\mathbb Z$ – Exodd May 20 '18 at 18:42
• Note: 1 should also say that $m$ and $n$ are relatively prime. If $m$ and $n$ are relatively prime then $n \mathbb{Z} + m \mathbb{Z} = \mathbb{Z}$ since there must exist integers $x,y$ such that $nx+my = 1$ (which is to say that $n\mathbb{Z}$ and $m\mathbb{Z}$ are comaximal). – user328442 May 20 '18 at 19:04
• Thank you so much! – Melissa Robles Carmona May 20 '18 at 19:41